I'm in trouble for understanding the normal bundle to $\mathbb S^n\subset\mathbb R^{n+1}$. By definition $$\nu(\mathbb S^n)=\bigcup_{p\in\mathbb S^n}\nu_p(\mathbb S^n),$$ where $\nu_p(\mathbb S^n)=\mathbb T_p\mathbb R^{n+1}/\mathbb T_p\mathbb S^n$. I have some questions about it:

(i) What do the elements of $\nu(\mathbb S^n)$ look like?

(ii) I know $T_\mathbb R^{n+1}\simeq \mathbb R^{n+1}$ whereas $T_p\mathbb S^n\simeq \mathbb R^n$, is it true that $T_p\mathbb R^{n+1}/\mathbb R^n\simeq \mathbb R^{n+1}/\mathbb R^n?$

(iii) Finally how can I stablish and isomorphism between $\nu(\mathbb S^n)$ and $\mathbb S^n\times \mathbb R$?



i) I think of elements as an ordered pair $(v,tv)$, where $|v|=1$ and $t$ is a real number. The normal bundle (when we're thinking about $S^n$ under its usual embedding in $\mathbb{R}^{n+1}$) is just the vectors that are orthogonal to $S^n$, so we can think of each point as lying on a line through the origin, so to nail down a a point we just need to know where this line pierces the sphere and how far along the line we are.

ii) I'm not sure what you mean by $\mathbb{R}^{n+1}/S^n$, I don't imagine that you mean that quotient space? If you meant to write $T_p \mathbb{R}^{n+1}/T_p S^n \cong \mathbb{R}^{n+1}/\mathbb{R}^n$, that is correct and sort of tautological.

iii) Do you know the result that a $1$ dimensional vector bundle is trivial if and only if it admits a nonvanishing section?

  • $\begingroup$ As to $(i)$ I can't use the definition of the normal bundle as the orthogonal vectors to $\mathbb S^n$, I must use the abstract definition of the quotient.. As to $(ii)$ I made a mistake when typing, the right is $\mathbb R^{n+1}/\mathbb R^n$ not $\mathbb R^{n+1}/\mathbb S^n$... As to $(iii)$ I know this result and I can use that... Thanks for the help.. $\endgroup$ – PtF Sep 24 '13 at 23:15
  • $\begingroup$ Then that isomorphism is correct (they are both simply isomorphic to $\mathbb{R}$). As for $i)$, you can still think about them as ordered pairs, the first coordinate just tells you where in that union you are, and then you just need to think about each fiber $T_p \mathbb{R}^{n+1}/T_p S^n$, try to grok how/why that's just a line. $\endgroup$ – James Cameron Sep 24 '13 at 23:20

(i) Normal bundle over $S^n$ looks like a porcupine rolled into a ball: there's an orthogonal needle (1-D line) sticking out of every point on the sphere.

(ii) It's better to picture $T_pS^n$ not as an abstract $R^n$, but as the tangent plane to $S^n$ at point $p$. What matters here is how $T_pS^n$ changes as you move $p$ along $S^n$.

(iii) Your assertion is wrong for even $n$. Here's how you can show that: the tautological sum of the tangent bundle and normal bundle would be the trivial bundle of $R^{n+1}$ over each point on $S^n$: you can identify it with the $R^{n+1}$ containing the sphere. For the proposed triviality of the normal bundle to hold the tangent bundle would need to be trivial as well. This is not the case for even $n$ because Euler characteristic of $S^n$ is non-zero. Alternatively, one can deny triviality of the tangent bundle by using fixed point theorem to show that even-dimensional spheres cannot admit non-vanishing vector fields.

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    $\begingroup$ I love your picture in $(i)$, but I think your (iii) is wrong. With vector bundles, one can have nontrivial bundle direct sum to a trivial bundle becoming trivial. In fact, spheres are the canonical example of this! The euler class (characteristic) is not a stable invariant, which is why your argument doesn't work. $\endgroup$ – Jason DeVito Sep 25 '13 at 0:00
  • $\begingroup$ Oops. Apparently my recollection of topology has a hole or two. :( It's so vague that I cannot even count the holes in it. :) You are right of course: "pointing outwards" would be a global section of the normal bundle, which makes it trivial. $\endgroup$ – Michael Sep 25 '13 at 16:28

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